Characterization of coextensive varieties of universal algebras
aa r X i v : . [ m a t h . C T ] A ug CHARACTERIZATION OF COEXTENSIVE VARIETIES OFUNIVERSAL ALGEBRAS
DAVID NEAL BROODRYK
Abstract.
A coextensive category can be defined as a category C with finite productssuch that for each pair X, Y of objects in C , the canonical functor × : X/ C ×
Y / C / / ( X × Y ) / C is an equivalence. We give a syntactical characterization of coextensive varietiesof universal algebras. An extensive category can be defined as a category C with finite coproducts such thatfor each pair X, Y of objects in C , the canonical functor + : C /X × C /Y / / C / ( X + Y )is an equivalence. A category that satisfies the dual condition is called coextensive.According to [1], the term “extensive category” was first used by W. F. Lawvere and S.Schanuel, although “categories with disjoint and universal coproducts” were consideredby A. Grothendieck a long time ago, and there are related papers of various authors.Examples of extensive categories include the category Cat of all small categories and thecategory
Top of topological spaces. An example of a coextensive category is the category
CRing of commutative rings.
1. Definition.
A category C with finite products is called coextensive if for each pair X, Y of objects in C , the canonical functor × : X/ C ×
Y / C / / ( X × Y ) / C is an equivalence.Equivalently let L be the left adjoint of × and let µ and ε be the unit and counit of thisadjunction, then L : ( X × Y ) / C / / X/ C ×
Y / C sends v : ( X × Y ) / / Z to the pair ofcanonical maps ( i , i ) : ( X, Y ) / / ( X + X × Y Z, Y + X × Y Z ) and C is coextensive if andonly if both ε and µ are natural isomorphisms. As follows from Proposition 2.2 of [1], C is coextensive if and only if for for everycommutative diagram A C p / / X × YA h (cid:15) (cid:15) X × Y Y π / / YC g (cid:15) (cid:15) AB p o o X × YX π o o XB f (cid:15) (cid:15) The two squares are pushouts if and only if the bottom row is a product diagram.Let C be a variety of universal algebras and let F ( X ) denote the free algebra in C on theset X . In particular we shall use F ( { x, y } ) which is the algebra consisting of terms of atmost two variables. We shall also use F ( ∅ ) which is the algebra consisting of all constantterms and is the initial object in C . (cid:13) David Neal Broodryk, 2020. Permission to copy for private use granted. DAVID NEAL BROODRYK
2. Proposition.
Any coextensive variety contains a ( k + 2) -ary term t and constants e , . . . , e k , e ′ , . . . , e ′ k ∈ F ( ∅ ) such that the identities hold: t ( x, y, e , . . . , e k ) = xt ( x, y, e ′ , . . . , e ′ k ) = y Proof.
Let A ⊆ F ( { x, y } ) × F ( { x, y } ) be the subalgebra generated by F ( ∅ ) × F ( ∅ ) ∪{ ( x, x ) , ( y, y ) } . Consider the diagram where i : A / / F ( { x, y } ) × F ( { x, y } ) is the inclusion: F ( ∅ ) × F ( ∅ ) A ⊂ (cid:15) (cid:15) F ( ∅ ) × F ( ∅ ) F ( ∅ ) π / / A F ( { x, y } ) p ◦ i / / AF ( { x, y } ) p ◦ i o o F ( ∅ ) × F ( ∅ ) F ( ∅ ) π o o F ( ∅ ) F ( { x, y } ) ⊂ (cid:15) (cid:15) F ( ∅ ) F ( { x, y } ) ⊂ (cid:15) (cid:15) AF ( { x, y } ) × F ( { x, y } ) i (cid:15) (cid:15) F ( { x, y } ) F ( { x, y } ) = (cid:15) (cid:15) F ( { x, y } ) F ( { x, y } ) = (cid:15) (cid:15) F ( { x, y } ) × F ( { x, y } ) F ( { x, y } ) p o o F ( { x, y } ) × F ( { x, y } ) F ( { x, y } ) p / / Note that since A contains the diagonal ∆( F ( { x, y } )) the maps p ◦ i and p ◦ i are sur-jective. On the other hand the elements of A are all of the form t (( x, x ) , ( y, y ) , ( e , e ′ ) , . . . , ( e k , e ′ k ))for some term t and constants e i , e ′ i . Therefore the quotients of A induced by p ◦ i and p ◦ i are exactly the quotients induced by π and π respectively. We conclude that thetop squares are pushout diagrams and therefore that the middle row is a product diagram.In particular ( x, y ) ∈ A , which implies:( x, y ) = t (( x, x ) , ( y, y ) , ( e , e ′ ) , . . . , ( e k , e ′ k )) = ( t ( x, y, e , . . . , e k ) , t ( x, y, e ′ , . . . , e ′ k ))for some term t and constants e i , e ′ i as desired.We call such an operation t a diagonalizing term and in any algebra A ⊇ F ( ∅ ) × F ( ∅ )we define the map δ ( x, y ) = t ( x, y, ( e , e ′ ) , . . . , ( e k , e ′ k )) HARACTERIZATION OF COEXTENSIVE VARIETIES OF UNIVERSAL ALGEBRAS
3. Proposition.
Let C be a variety of universal algebras containing a diagonalizing term t equipped with constants e , . . . , e k , e ′ , . . . , e ′ k . Then the following conditions hold:1. C is left coextensive2. for any X, Y ∈ C and any surjective q : X × Y / / Z there exist surjective f : X / / X ′ and g : Y / / Y ′ such that q ≃ f × g . Proof. C is left coextensive by Theorem 3 of [2] since we can fix any constant as andhave the equalities: t ( x, , e , . . . , e k ) = xt ( x, , e ′ , . . . , e ′ k ) = 0 Let q : X × Y / / Z be surjective. We define the relations: • E X = { ( a, c ) ∈ X × X | ∃ b,d ∈ Y : q (( a, b )) = q (( c, d )) }• E Y = { ( b, d ) ∈ Y × Y | ∃ a,c ∈ X : q (( a, b )) = q (( c, d )) } In any variety these relations are reflexive, symmetric and homomorphic. To showthat they are also transitive let ( a, c ) , ( c, f ) ∈ E X . Then there exist b, d, e, g such that q (( a, b )) = q (( c, d )) and q (( c, e )) = q (( f, g )) . q (( a, e )) = q ( t (( a, b ) , ( c, e ) , ( e , e ′ ) , . . . , ( e k , e ′ k )))= q ( t (( c, d ) , ( c, e ) , ( e , e ′ ) , . . . , ( e k , e ′ k )))= q (( c, e ))= q (( f, g )) Therefore E X and E Y are congruences and we can define the quotient projections: q X : X / / X/E X , q Y : Y / / Y /E Y , and q x × q y : X × Y / / ( X × Y ) / ( E x × E y ) . Clearlyif q (( a, b )) = q (( c, d )) then ( q x × q y )(( a, b )) = ( q x × q y )(( c, d )) . On the other hand, if ( q x × q y )(( a, b )) = ( q x × q y )(( c, d )) then there exist a ′ , c ′ ∈ Y and b ′ , d ′ ∈ X such that q (( a, a ′ )) = q (( c, c ′ )) and q (( b ′ , b )) = q (( d ′ , d )) Therefore we have: q ( a, b ) = q ( t (( a, a ′ ) , ( b ′ , b ) , ( e , e ′ ) , . . . , ( e k , e ′ k )))= q ( t (( c, c ′ ) , ( d ′ , d ) , ( e , e ′ ) , . . . , ( e k , e ′ k )))= q ( c, d ) So q ≃ q x × q y as desired. DAVID NEAL BROODRYK
4. Proposition.
Let C be a variety, the following statements are equivalent:1. C is coextensive2. C contains a diagonalizing term and for any A ∈ C and any equalities in F U ( A ) : u ( a , . . . , a m , ω , . . . , ω n ) = v ( a , . . . , a m , ω , . . . , ω n ) u ( a , . . . , a m , ω ′ , . . . , ω ′ n ) = v ( a , . . . , a m , ω ′ , . . . , ω ′ n ) we have the equality in F U ( A ) + F ( ∅ ) : u ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ )) = v ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ )) C contains a diagonalizing term and for any set X satisfies the equations:(a) δ ( x, x ) = x for any x ∈ F ( X ) + F ( ∅ ) (b) u ( δ ( x , y ) , . . . , δ ( x n , y n )) = δ ( u ( x , . . . , x n ) , u ( y , . . . , y n )) for any operation u and any x i , y i ∈ F ( X ) + F ( ∅ ) Proof. (1 ⇐⇒
2) If C is coextensive then by Proposition 2, C contains a diagonalizingterm t . On the other hand if C contains such a term then by Proposition 3.1, C is leftcoextensive. It remains to show that in any such C , when diagram 1 is a pushout diagramit is also a product diagram. It is sufficient to consider only X = Y = F ( ∅ ). In particularconsider for any algebra A equipped with a morphism h : F ( ∅ ) / / A the followingpushout diagram: F ( ∅ ) × F ( ∅ ) F U ( A ) + F ( ∅ ) ⊂ (cid:15) (cid:15) F ( ∅ ) × F ( ∅ ) F ( ∅ ) π / / F U ( A ) + F ( ∅ ) F U ( A ) p o o F U ( A ) + F ( ∅ ) F U ( A ) p / / F ( ∅ ) × F ( ∅ ) F ( ∅ ) π o o F ( ∅ ) F U ( A ) ⊂ (cid:15) (cid:15) F ( ∅ ) F U ( A ) ⊂ (cid:15) (cid:15) F U ( A ) + F ( ∅ ) A [ ε A ,h ] (cid:15) (cid:15) F U ( A ) B f (cid:15) (cid:15) F U ( A ) C g (cid:15) (cid:15) AB p ′ o o A C p ′ / / HARACTERIZATION OF COEXTENSIVE VARIETIES OF UNIVERSAL ALGEBRAS U is the underlying set functor and + denotes the coproduct. All four smallsquares are defined to be pushouts. In particular note that [ ε A , h ] is surjective, so if thefirst row is a product diagram, then by Proposition 3.2 so is the second row and the wholesquare. Therefore it is sufficient to consider only the top row.Consider the map ( p , p ) : F U ( A ) + F ( ∅ ) / / F U ( A ) . C will be coextensive if andonly if this map is a bijection. Note that for any a, b ∈ F U ( A ) we have:( p , p )( δ ( a, b )) = ( t ( a, b, e , . . . , e k ) , t ( a, b, e ′ , . . . , e ′ k )) = ( a, b )and so ( p , p ) is surjective. On the other hand it is injective if and only if for all pairsof identities: u ( a , . . . , a m , ω , . . . , ω n ) = v ( a , . . . , a m , ω , . . . , ω n ) u ( a , . . . , a m , ω ′ , . . . , ω ′ n ) = v ( a , . . . , a m , ω ′ , . . . , ω ′ n )we have: u ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ )) = v ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ ))for some operations u, v , variables a , . . . , a m , and constants ω , . . . , ω n , ω ′ , . . . , ω ′ n .(2 ⇐⇒
3) Let X be any set and note that if 2 holds we immediately have:1. δ ( x, x ) = x for any x ∈ F ( X ) + F ( ∅ ) u ( δ ( x , y ) , . . . , δ ( x n , y n )) = δ ( u ( x , . . . , x n ) , u ( y , . . . , y n )) for any operation u andany x i , y i ∈ F ( X ) + F ( ∅ ) On the other hand if these two equations hold then for any such pair of identities wehave: u ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ )) = u ( δ ( a , a ) , . . . , δ ( a m , a m ) , δ ( ω , ω ′ ) , . . . , δ ( ω n , ω ′ n ))= δ ( u ( a , . . . , a m , ω , . . . , ω n ) , u ( a , . . . , a m , ω ′ , . . . , ω ′ n ))= δ ( v ( a , . . . , a m , ω , . . . , ω n ) , v ( a , . . . , a m , ω ′ , . . . , ω ′ n ))= v ( δ ( a , a ) , . . . , δ ( a m , a m ) , δ ( ω , ω ′ ) , . . . , δ ( ω n , ω ′ n ))= v ( a , . . . , a m , ( ω , ω ′ ) , . . . , ( ω n , ω ′ )) DAVID NEAL BROODRYK
5. Theorem.
A variety C is coextensive if and only if the following conditions hold:1. C contains a ( k + 2) -ary term t and constants e , . . . , e k , e ′ , . . . , e ′ k ∈ F ( ∅ ) such thatthe identities hold: t ( x, y, e , . . . , e k ) = xt ( x, y, e ′ , . . . , e ′ k ) = y
2. There exist terms α , . . . , α n , β , . . . , β n ∈ F ( { x } + U ( F ( ∅ ) )) , operations u , . . . , u m ∈ F ( X ) and for all operations s ∈ F ( X ) there exist u ( s )0 , . . . , u ( s ) m such that u ( β , . . . , β n ) = δ ( x, x ) u m ( α , . . . , α n ) = xu ( s )0 ( β , . . . , β n ) = δ ( s ( x , . . . , x l ) , s ( x ′ , . . . , x ′ l )) u ( s ) m ( α , . . . , α n ) = s ( δ ( x , x ′ ) , . . . , δ ( x l , x ′ l )) and for all i < m : u i ( α , . . . , α n ) = u i ( β , . . . , β n ) u ( s ) i ( α , . . . , α n ) = u ( s ) i ( β , . . . , β n )
3. For each j < n one of the following is true:(a) α j = β j (b) α j = ( ω, ω ′ ) and β j = δ ( ω, ω ′ ) for some constants ω, ω ′ (c) β j = ( ω, ω ′ ) and α j = δ ( ω, ω ′ ) for some constants ω, ω ′ (d) α j = v ( δ ( ω , ω ′ ) , . . . , δ ( ω l , ω ′ l )) and β j = δ ( v ( ω , . . . , ω l ) , v ( ω ′ , . . . , ω ′ l )) for someconstants ω , . . . , ω l , ω , . . . , ω ′ l (e) β j = v ( δ ( ω , ω ′ ) , . . . , δ ( ω l , ω ′ l )) and α j = δ ( v ( ω , . . . , ω l ) , v ( ω ′ , . . . , ω ′ l )) for someconstants ω , . . . , ω l , ω , . . . , ω ′ l where delta ( a, b ) = t ( a, b, ( e , e ′ ) , . . . , ( e k , e ′ k )) and X is any set. HARACTERIZATION OF COEXTENSIVE VARIETIES OF UNIVERSAL ALGEBRAS Proof.
Condition 1 simply says that C has a diagonalizing term. Let 1 F ( X ) + ε F ( ∅ ) : F ( X ) + F U ( F ( ∅ ) ) / / F ( X ) + F ( ∅ ) be the map induced by the identity at F ( X ) and thecounit at F ( ∅ ). Then F ( X ) + F ( ∅ ) ≃ F ( X ∪ U ( F ( ∅ ) )) /E where E is the congruencegenerated by this map. Note that this congruence is generated by the relation R = { δ ( ω, ω ′ ) = ( ω, ω ′ ) } ∪ { u ( δ ( ω , ω ′ ) , . . . , δ ( ω n , ω ′ n )) = δ ( u ( ω , . . . , ω n ) , u ( ω ′ , . . . , ω ′ n )) } Note that δ ( x, x ) = x for any x ∈ F ( X ) + F ( ∅ ) if and only if ( δ ( x, x ) , x ) ∈ E ifand only if there exists a natural number n such that ( δ ( x, x ) , x ) ∈ Q n , where Q is thereflexive symmetric homomorphic relation on F ( X ) + F U ( F ( ∅ ) ) generated by R . Thisis true if and only if there exist a , . . . , a k ∈ F ( X ) + F U ( F ( ∅ ) ) such that a = δ ( x, x ), a k = x and ( a i , a i +1 ) ∈ Q for i < k . But ( a i , a i +1 ) ∈ Q if and only if for some term u i wehave: a i = u i ( β , . . . , β n ) a i +1 = u i ( α , . . . , α n )Where the α j and β j are in the symmetric reflexive relation generated by R . Thisis true exactly when condition 3 above is satisfied. Similarly for any operation s andvariables x , . . . , x n , x ′ , . . . , x ′ n ∈ X there exist a ′ , . . . , a ′ k ∈ F ( X ) + F U ( F ( ∅ ) ) such that a ′ = s ( δ ( x , x ′ ) , . . . , δ ( x n , x ′ n )), a ′ k = δ ( s ( x , . . . , x n ) , s ( x ′ , . . . , x ′ n )) and for i < k thereexists some term u ′ i such that: a i = u ( s ) i ( β , . . . , β n ) a i +1 = u ( s ) i ( α , . . . , α n )Where the α j and β j are in the symmetric reflexive relation generated by R . Tosimplify the notation we can assume that the α j and β j are the same for every u i and u ( s ) i .Now by proposition 4, we have that C is coextensive if and only if the above conditionshold.Recall that δ ( x, y ) is not an operation in the usual sense but is instead defined as δ ( x, y ) = t ( x, y, ( e , e ′ ) , . . . , ( e k , e ′ k )) Therefore equations involving δ are actually equa-tions involving t and the elements ( e , e ′ ) , . . . , ( e k , e ′ k ). Note that F ( X ) + F U ( F ( ∅ ) ) = F ( X + U ( F ( ∅ ) )) is a free algebra in which the ( ω, ω ′ ) are elements of the generating set.Therefore the equations of condition 2 in the theorem must still hold after substituting( ω, ω ′ ) = x ω,ω ′ for some variables x ω,ω ′ ∈ X . As an example it is helpful to consider thevariety of commutative rings. DAVID NEAL BROODRYK
6. Example.
Let C be the variety of commutative rings. let t ( a, b, c, d ) = a · c + b · d , e = 1 , e = 0 , e ′ = 0 , e ′ = 1. Then we have t ( a, b, e , e ) = a · b · a and t ( a, b, e ′ , e ′ ) = a · b · b , so t is a diagonalizing term.To show δ ( x, x ) = x consider: u ( a , a , a , a ) = a · a + a · a u ( a , a , a , a ) = a · a α = (cid:0) x, (1 , , (0 , , (cid:1) β = (cid:0) x, (1 , , (0 , , · (1 ,
0) + 1 · (0 , (cid:1) Note in particular that α = 1 and β = 1 · (1 ,
0) + 1 · (0 ,
1) satisfies condition 3( d ) ofthe theorem. This gives the equations: δ ( x, x ) = x · (1 ,
0) + x · (0 ,
1) = x · (1 · (1 ,
0) + 1 · (0 , ≃ E x · x Therefore δ ( x, x ) = x as desired. It remains to show that s ( δ ( x , y ) , . . . , δ ( x n , y n )) = δ ( s ( x , . . . , x n ) , u ( y , . . . , y n )) for any operation s and any x i , y i ∈ F U ( A ) + F ( ∅ ) . Itsuffices to show this for s ( x, y ) = x + y and s ( x, y ) = x · y .To show δ ( x + x ′ , y + y ′ ) = δ ( x, y ) + δ ( x ′ , y ′ ) consider: u ( a , a , a , a , a , a ) = a · a + a · a + a · a + a · a α = (cid:0) x, x ′ , y, y ′ , (1 , , (0 , (cid:1) β = (cid:0) x, x ′ , y, y ′ , (1 , , (0 , (cid:1) Note that in this case no equivalence under E is necessary so we have α = β . Thisgives the equations: δ ( x + x ′ , y + y ′ ) = ( x + x ′ ) · (1 , y + y ′ ) · (0 ,
1) = x · (1 , y · (0 , x ′ · (1 , y ′ · (0 ,
1) = δ ( x, y )+ δ ( x ′ , y ′ )To show δ ( x · x ′ , y · y ′ ) = δ ( x, y ) · δ ( x ′ , y ′ ) consider: u ( a , . . . , a
1) = a a a + a a a + a a a + a a a u ( a , . . . , a
1) = a a a + a a a + a a a + a a a u ( a , . . . , a
1) = ( a a a a · ( a a a a α = (cid:0) x, x ′ , y, y ′ , (1 , , (0 , , , δ (1 · , · , δ (1 · , · , δ (0 · , · , δ (1 , , δ (0 , (cid:1) β = (cid:0) x, x ′ , y, y ′ , δ (1 , , δ (0 , , δ (0 , , δ (1 , , δ (1 , δ (0 , , δ (0 , , (1 , , (0 , (cid:1) This gives the equations:
HARACTERIZATION OF COEXTENSIVE VARIETIES OF UNIVERSAL ALGEBRAS δ ( x · x ′ , y · y ′ ) = xx ′ · (1 ,
0) + yy ′ · (0 , xx ′ · (1 ,
0) + xy ′ · yx ′ · yy ′ · (0 , ≃ E xx ′ · δ (1 ,
0) + xy ′ · δ (0 ,
0) + yx ′ · δ (0 ,
0) + yy ′ · δ (0 , xx ′ · δ (1 · , ·
0) + xy ′ · δ (1 · , ·
1) + yx ′ · δ (1 · , ·
1) + yy ′ · δ (0 · , · ≃ E xx ′ · δ (1 , + xy ′ · δ (1 , δ (0 ,
1) + yx ′ · δ (1 , δ (0 ,
1) + yy ′ · δ (0 , = ( x · δ (1 ,
0) + y · δ (0 , · ( x ′ · δ (1 ,
0) + y ′ · δ (0 , ≃ E ( x · (1 ,
0) + y · (0 , · ( x ′ · (1 ,
0) + y ′ · (0 , δ ( x, y ) · δ ( x ′ , y ′ )And so the variety of commutative rings is coextensive. References [1] A. Carboni, S. Lack, and R. F. C. Walters, Introduction to extensive and distributivecategories, Journal of Pure and Applied Algebra, 84(2), 1993, 145-158.[2] D. Broodryk, Characterization of left coextensive varieties of universal algebras, Theoryand Applications of Categories, Vol. 34, No. 32, 2019, 1036-1038.
Department of Mathematics and Applied Mathematics, University of Cape TownRondebosch 7701